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I have been learning recently about elliptic curves, and I am totally new in this topic. I just followed the construction of the group-scheme structure on elliptic curves given by Vakil in his notes (that you can find here). Right after this, he gives a pretty concrete exercise (page 522) for which I would like some hints as to how to solve it efficiently.

Here is the statement of the exercise.

Let $k$ be a field. Consider the genus $1$ curve $C\subset \mathbb{P}^2_k$ given by $$y^2z=x^3+x^2z$$ with the point $\mathbb{p}=[0,1,0]$. Emulate "the above argument" to show that $C \setminus \{[0,0,1]\}$ is a group variety. Show that it is isomorphic to $\mathbb{G}_m$ (the multiplicative group-scheme $\operatorname{Spec}(k[t,t^{-1}])$) with coordinate $t=\frac{y}{x}$ by giving an isomorphism of schemes, and showing that multiplication and inverse in both group-varieties agree under this isomorphism.

Okay, there are multiple things going on here. I am going to ask a few questions about this. Some of them may be pretty basic, but I think that this exercise would is a perfect occasion to clear up a few confusions in my mind about basic computations in algebraic geometry. In the following, let $U$ denote $C \setminus \{[0,0,1]\}$.

First, to justify that $U$ has a natural structure of scheme, I prove that $[0,0,1]$ is a closed point by computing its residue field in the distinguished open set $z\not = 0$. Here, $C$ becomes the affine curve $y^2=x^3+x^2$ in $\mathbb{A}^2_k=\operatorname{Spec}(k[x,y])$ and the point we are interested in is $(0,0)$. Its residue field is the quotient field of the integral domain $A/(x,y)$, where $A=k[x,y]/(y^2-x^2-x^3)$. This integral domain actually already is a field, none other than $k$. As a result, $[0,0,1]$ is a closed point of $C$ of degree $1$.
Is this method the correct way to proceed? Is there a more straightforward way to see it?

Second, to show that $U$ comes with a structure of group-variety, I actually need to show that it is a subgroup-scheme of $C$, ie that it is stable under the group laws and contains the neutral point $\mathbb{p}$. Actually, I do not really understand which "above argument" Vakil is refering to here. From what I can tell, I have to check by hand that whenever I consider a point in $U$ and look at the line joining it to $\mathbb{p}$, then the third point of intersection with $C$ can not be $[0,0,1]$ ; and the same goes whenever I consider the line joining two arbitrary points of $U$.
I think I have actually been able to show the first part, but this is quite fastidious and I feel like there must be another way to check this properly. How would you treat this part of the exercise?

Third, to build the scheme-theoretic morphism between $U$ and $\mathbb{G}_m$, it is enough to give a $k$-linear morphism $k[t,t^{-1}]\rightarrow \mathcal{O}_{U}(U)$. Now, I notice that $U$ is just the union of the distinguished open sets $x\not = 0$ and $y \not = 0$, intersected with $C$. The global sections of $U$ are the pairs of global sections on each of these open sets that agree on the intersection. Global sections on each of the open sets are given by a quotient of polynomials whose denominator is non-vanishing. So I see that I could send $t$ to the pair $(\frac{y}{x},\frac{x}{y})$, but these do not agree on the intersection as far as I can tell.
How does one makes sense of the statement $t=\frac{y}{x}$ given in the exercise as a hint?

Fourth, I know as a general result that a proper geometrically connected curve over a field becomes affine when a finite nonzero number of points are omitted from it (see here). Hence, it is not surprizing to see that it applies for $U$. However, is there a way to compute this affine space easily in conrete examples? Just by seeing the equation and the point we are omitting, is there a way to guess that we will obtain something nice like $\operatorname{Spec}(k[t,t^{-1}])$? How does one infer what the ring must look like?

Thank you very much for reading through this. Any clarification for any of the above points will be greatly appreciated.

Edit: As pointed out in the comment, $C$ is not a group-scheme, hence my second point reasoning makes no sense. Let me then replace my second point with the following: in order to show that $U$ is a group-scheme, can I simply argue that $U$ is an elliptic curve, because it is obtained by removing the singular point(s) of a genus $1$ projective curve described by a Weierstraß type equation? If so, this part of the exercise now becomes trivial. What is important is to understand why the group-structure behaves like $\mathbb{G}_m$.

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  • $\begingroup$ Your second point is not correct. $C$ is NOT a group scheme, only $U$ is. May be the easiest path is to show that the normalization of $C$ is $\mathbb{P}^1$ and the fiber over $[0,0.1]$ is two points and outside it is an isomorphism. Then, $U$ is just the projective line minus two points ($k$-rational) which has the structure of the multiplicative group scheme. $\endgroup$ – Mohan Jul 5 '18 at 0:16
  • $\begingroup$ @Mohan So $C$ is not a group-scheme? I am a little confused. The data of $C$ together with the $k$-point $\mathbb{p}$ is an elliptic curve, right? Why is there no group-scheme structure defined on it? (Would it be because $[0,0,1]$ is not regular?) $\endgroup$ – Suzet Jul 5 '18 at 0:53
  • $\begingroup$ Ok, I have convinced myself that it is indeed not regular, hence one would have to remove it in order to get an elliptic curve. So I see that any genus $1$ curve described by the Weierstraß normal form is not necessarily an elliptic curve as a whole - one may need to remove singular points for this. I was absolutely not aware of this fact. $\endgroup$ – Suzet Jul 5 '18 at 1:07
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    $\begingroup$ A singular cubic is never an elliptic curve. Even if you remove the singular point, you only get a rational curve, never an elliptic curve even birationally. $\endgroup$ – Mohan Jul 5 '18 at 3:26
  • $\begingroup$ I see. I am going to ask another question about the Weierstraß form then, because I feel a little confused about it. My misunderstanding may come from this confusion. $\endgroup$ – Suzet Jul 5 '18 at 4:25

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